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JEE Main 2024
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Medium

Question

 If y=cos(π3+cos1x2), then (xy)2+3y2 is equal to \text { If } y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \text {, then }(x-y)^2+3 y^2 \text { is equal to }

Answer: 3

Solution

y=cos(π3+cos1x2)=cos(π3)cos(cos1(x2))sin(π3.)sin(cos1(x2))=12x2321x244y=x34x2(4yx)2=3(4x2)16y2+x28xy=123x2x2+4y22xy=3(xy)2+3y2=3\begin{aligned} & y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \\ & =\cos \left(\frac{\pi}{3}\right) \cos \left(\cos ^{-1}\left(\frac{x}{2}\right)\right)-\sin \left(\frac{\pi}{3 .}\right) \sin \left(\cos ^{-1}\left(\frac{x}{2}\right)\right) \\ & =\frac{1}{2} \cdot \frac{x}{2}-\frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^2}{4}} \\ & \Rightarrow 4 y=x-\sqrt{3} \sqrt{4-x^2} \\ & \Rightarrow(4 y-x)^2=3\left(4-x^2\right) \\ & \Rightarrow 16 y^2+x^2-8 x y=12-3 x^2 \\ & x^2+4 y^2-2 x y=3 \\ & (x-y)^2+3 y^2=3 \end{aligned}

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